3-1
Solutions for Chapter 3 – HW4 Problems
P3.27: For the coaxial cable example 3.8, we found:
Ir
for r £ a, H= af ,
2p a 2
I
for a<r £ b, H = af ,
2pr
I æ c2 - r 2 ö
for b<r £ c, H= ç ÷ af ,
( )
2p c 2 - b 2 è r ø
and for c<r , H =0.
a. Evaluate the curl in all 4 regions.
b. Calculate the current density in the conductive regions by dividing the current by
the area. Are these results the same as what you found in (a)?
1 ¶ æ Ir2 ö I
(a) Ñ ´ H = ç a = 2 a z for r £ a
2 ÷ z
r ¶r è 2p a ø pa
1 ¶ æ I ö
Ñ´H = a = 0 for a < r < b
r ¶r çè 2p ÷ø z
1 ¶ æ I (c - r ) ö
2 2
-I
Ñ´H = ç ÷a = a z for b<r £ c
r ¶r 2p ( c - b )
ç 2 2 ÷ z
p ( c 2
- b 2
)
è ø
Ñ ´ H = 0 for r > c
( )
(b) for r £ a, S = p a 2 , J = I a z = 2 a z
S
I
pa
-I
for b<r £ c, S = p ( c 2 - b 2 ) , J = az
p ( c2 - b2 )
Comment: Ñ ´ H = J is confirmed.
3-2
P3.28: Suppose you have the field H = r cos q af A/m. Now consider the cone specified
by q = p/4, with a height a as shown in Figure 3.56. The circular top of the cone has a
radius a.
a. Evaluate the right side of Stoke’s theorem through the dS = dSaq surface.
b. Evaluate the left side of Stoke’s theorem by integrating around the loop.
1 é ¶ ù 1 ¶ ( rH f )
(a) Ñ ´ H = ê
r sin q ë ¶q
( sin q H f ) ú a r -
û r ¶r
aq
ar derivative:
¶ 1 é ¶ ù ( cos 2 q - sin 2 q )
( r sin q cos q ) = r ( cos 2 q - sin 2 q ) ; êë ¶q ( sin q H f ) úû a r = ar
¶q r sin q sin q
Fig. P3.28
aq derivative:
-1 ¶ ( r cos q )
2
aq = -2 cos q aq
r ¶r
So, Ñ ´ H =
( cos2 q - sin 2 q ) a - 2 cos q a
q
sin q
r
Now we must integrate this over the aq surface:
é ( cos 2 q - sin 2 q ) ù
ò Ñ ´ H dS = òê
ê a - 2 cos q aq ú r sin q drdf ( -aq )
sin q
r
ë úû
2a 2p
= ò 2r sin q cos q drdf =2 ( sin q cos q ) q = p ò rdr ò df = 2p a
2
4 0 0
2p
ò H dL = ò r r =a cos q af adf af = a 2 cos q q = p ò df = 2p a
2
(b) 2
4 0
Clearly in this case the circulation of H is the easiest approach.
3-3
5. Magnetic Flux Density
P3.31: An infinite length coaxial cable exists along the z-axis, with an inner shell of radius
a carrying current I in the +az direction and outer shell of radius b carrying the return
current. Find the magnetic flux passing through an area of length h along the z-axis
bounded by radius between a and b.
I µo I
H= af , B = af ,
2pr 2pr
µo I af µI
d r dzaf = o ln ( r ) a h
b
For a < r < b, f = ò B dS = ò
2p r 2p
µo Ih æ b ö
f= ln ç ÷ Wb
2p èaø
3-4
6. Magnetic Forces
P3.34: A 10. nC charged particle has a velocity v = 3.0ax + 4.0ay + 5.0az m/sec as it enters
a magnetic field B = 1000. T ay (recall that a tesla T = Wb/m2). Calculate the force vector
on the charge.
é m Wb ù
F = q ( u ´ B ) = 10 x10-9 C ê( 3a x + 4a y + 5a z ) ´1000 2 a y ú
ë s m û
The cross-product:
éa x a y az ù
ê3 4 5 úú = -5000a x + 3000a z
ê
êë 0 1000 0 úû
Evaluating we find: F = -50ax + 30az µN
P3.40: Suppose you have a pair of parallel lines each with a mass per unit length of 0.10
kg/m. One line sits on the ground and conducts 200. A in the +ax direction, and the other
one, 1.0 cm above the first (and parallel), has sufficient current to levitate. Determine the
current and its direction for line 2.
F12 µo I1 I 2 4p x10-7 H m 200 A
Here we will use = ay = I 2a z = 4 x10-3 I 2a z
L 2p y 2p 0.01m
2
F mg 2 Ns N
= = ( 0.10 kg m ) ( 9.8 m s ) = 0.98
L L kgm m
So solving for I2:
0.98
I2 = = 245 A in the -a x direction.
4 x10-3
3-5
P3.41: In Figure 3.57, a 2.0 A line of current is shown on the z-axis with the current in the
+az direction. A current loop exists on the x-y plane (z = 0) that has 4 wires (labeled 1
through 4) and carries 1.0 mA as shown. Find the force on each arm and the total force
acting on the loop from the field of the 2.0 A line.
µo I1
F12 = ò I 2 dL2 ´ B1 ; B1 = af
2pr
A ® B : dL2 = 5df af , af ´ af = 0
C ® D : af ´ af = 0
3
µo I1 µ I I æ3ö
B ® C : F12 = I 2 ò d r a r ´ af = o 1 2 ln ç ÷ a z
5
2pr 2p è5ø
Fig. P3.41
4p x10-7 H m
( 2 A) (10-3 A) ln æç ö÷ a z = -204 pNa z
3
F12 =
2p è5ø
So for B to C: F12 = -0.20 nN az
Likewise, from D to A: F12 = +0.20 nN az
